Functions & Relations Question - Need help...

Hey guys, I was hoping someone could explain how to answer this question as I am pretty much clueless. Any help is appreciated!


Consider the relation on the set of integers Z defined by:

aRb if ab>0 or a=b=0

a. Show that this is an equivalence relation.
b. Find the equivalence classes of this relation.


This question is copied exactly from the past paper.

Thanks!

For part a:
What three properties does a relation need to have for it to be an equivalence relation? Can you show that the relation R satisfies these three properties?

For part b:
In general, there are lots of ways of finding the equivalence classes - remember an equivalence class (in this case) is just a subset of Z such that if we pick any two elements x and y in that subset, we have xRy. One possible way of finding the equivalence classes is to pick an element of Z, and find all the elements which are in the same equivalence class as it. Now, pick another element of Z, and repeat this procedure. Keep doing this until all the elements of Z are in some equivalence class.

For part a: Does that mean i have to show that it satisfy the three axioms? If i do so, does that prove that its an equivalence relation? And i am still a little confused about part b.

Can you please write out your answer for this question?

Thank you for the help btw!

For part a, the definition of an equivalence relation is a relation that satisfies the three axioms you're talking about. So, if you can show that the relation does satisfy the three axioms, then it must be an equivalence relation.

To show them as equivalence relations through the three axioms, would these be correct as an answer?

For ab>0

Associativity: (a) x b = a x (b) = 1 x 2 = 2 x 1 (a = 1, b = 2)

Identity: a x e = e x a = e = 1 x 1 = 1 x 1 = 1 (e = 1)

Inverse: a x a^-1 = a^-1 x a = e = 1 x 1 = 1 x 1 = 1 (a^-1 = 1)

For a = b = 0

Associativity: (a) x b = a x (b) = 0 x 0 = 0 x 0 (a = b = 0)

Identity: a x e = e x a = e = 0 x 0 = 0 x 0 = 0 (e = 0)

Inverse: a x a^-1 = a^-1 x a = e = 0 x 0 = 0 x 0 = 0 (a^-1 = 0)


I'm sorry about the late reply, because your help has been excellent! So glad that Sky isn't my broadband provider anymore!

Those are three of the four group axioms -- you're working with a relation here, not a group. A relation is an equivalence relation if it is reflexive, symmetric and transitive -- these are the three things you need to check. [I'm not even really sure what you were doing with the group axioms... for example, unless a=1 or a=-1, $a^{-1}$ isn't even an integer. But this doesn't matter since group theory has nothing to do with this problem anyway.]

So for your question, to do part (a) you need to check that the relation is reflexive, symmetric and transitive. Then to do part (b), you need to think of what might constitute an equivalence class. The hint is really in the >0 bit... if a>0, what is [a] (the equivalence class of a)? If a<0, what is [a]? What is [0]? Are there any more equivalence classes?

Ok. I think I went off track a little... Could you show me how you would show it is an equivalence relation? Or use a similar example, because I have my notes here and they are not really helping.

I'll have a little guess:

~ is a reflexive relation if it satisfies the condition:

a ~ a , for any element a in Z

a=b = a=b so then it is true? :/

Hey guys, I was hoping someone could explain how to answer this question as I am pretty much clueless. Any help is appreciated!


Consider the relation on the set of integers Z defined by:

aRb if ab>0 or a=b=0

a. Show that this is an equivalence relation.
b. Find the equivalence classes of this relation.


This question is copied exactly from the past paper.

Thanks!

In general, a relation R is:
- reflexive if, for all a, aRa
- symmetric if bRa whenever aRb (i.e. aRb implies bRa)
- transitive if, whenever aRb and bRc, aRc

Take for example the relation $\le$ on $\mathbb{Z}$ (i.e. aRb if $a \le b$; we just write $\le$ instead of R). This is reflexive, since for any integer a, it's definitely true that $a \le a$. It's also transitive, since if $a \le b$ and $b \le c$ then $a \le c$. But it's not symmetric, since if $a<b$ then $a \le b$, but it's not true that $b \le a$ in this case.

In your problem, aRb if ab>0 or a=b=0. You need to show that it's reflexive, symmetric and transitive.

If so, then ab>0 can't be an equivalence relation can it? Because the symmetric axiom would say that if a>b then b>a which can't be true.

No... what does a>b and b>a have to do with the statement ab>0?

Like, it is an equivalence relation; that's why you're being asked to show that it's an equivalence relation.

Is the part where I showed that a=b=0 is an equivalence relation at least correct?

That bit's fine, yup. You now need to consider the case where the elements are nonzero, using the fact that a~b if ab>0. Is it true that a~a? If a~b, is it true that b~a? If a~b and b~c, is it also true that a~c? [For each of these questions, convert the "a~b" (etc) parts into mathematical inequalities using the definition of ~.]

Ah, ok. So would I, for example, show the reflexive relation as:

a>0 = a>0 ?

It shows that it is positive but doesn't quite look correct.

Also, would the equivalence classes be: The positive integers, negative integers and zero then? < would that be acceptable as an answer even?

It's not correct. What is the = sign there for? What is the second element being multiplied on the LHS of the inequality? Etc. For reflexivity, you know that a~b means that ab>0. So what does a~a mean? Does this always hold when a is nonzero? Use a similar method for the other two.

If you do this right, you shouldn't need any = signs anywhere.


That's correct, but you need to give justification for the claim.

Reflexive: aa>0 is always true if a>0

Symmetric: if ab>0 then ba>0 must be true

Transitive: if ab>0 and bc>0 then ac>0 must be true

And as for the equivalence classes, how can I show justification? Mathematically or worded?

You need to show that this is true for all $a \ne 0$, not just $a>0$. How else can you write aa?

Yup... but why? [It's extremely obvious, but it's worth writing it anyway.]

Again this is true, but you need to say why (without assuming anything about the positivity or negativity of a, b or c). This one is less obvious than the previous one, so it's definitely worth saying why it must be true.

Well, you should show it mathematically, but it can be worded and mathematical at the same time. Given $a$, the equivalence class $[a]$ is the set of real numbers $b$ such that $ab>0$. For the three different cases of $a$ that you identified above, derive $[a]$ from this definition.

Huh? I mean how else can you write $aa$? Or, to make it more obvious, how else can you write $a \times a$? You can't assume that $a>0$, you can only assume that $a \ne 0$ (so $a$ could be negative). But it's still true that $a~a$. You need to say why.

No; like I say, you can't assume that $a$ or $b$ is positive. But you've said that if $ab>0$ then $ba>0$. You just need to say why this is true.

This still isn't the right reasoning. I'll give you a hint: you're assuming that $ab>0$ and $bc>0$, which means that when you multiply $ab$ and $bc$ together, this is also greater than zero. What can you deduce from this?